Exercise 13.2.11

Proof.

$\text{∙}$

Suppose that ${𝜃}_f(y)$ has no root in $F$ (lines 1 and 2 of the table). By Theorem 13.2.6 (b), $G$ is not conjugate to a subgroup of $\mathrm{AGL}(1,{𝔽}_5)$. Therefore by diagram (13.6) and Theorem 13.2.2, $G=A_5$ or $G=S_5$ (no conjugacy here). By Theorem 13.2.6 (a), $G=A_5$ if $\mathrm{\Delta }(f)\in F^2$, and $G=S_5$ otherwise.

$\text{∙}$

Suppose now that ${𝜃}_f(y)$ has a root in $F$ (lines 3,4,5 of the table). Then, by Theorem 13.2.6 (b) (and Theorem 13.2.2), $G$ is conjugate to a subgroup of $\mathrm{AGL}(1,{𝔽}_5)$ containing $⟨(12345)⟩$.

So, by diagram (13.16), $G$ is conjugate to $\mathrm{AGL}(1,{𝔽}_5),\mathrm{AGL}(1,{𝔽}_5)\cap A_5$, or $⟨(12345)⟩$.

If $\mathrm{\Delta }(f)\notin F^2$, $G\not\subset A_5$, therefore $G=\mathrm{AGL}(1,{𝔽}_5)$. This is the third line of the table.

If $\mathrm{\Delta }(f)\in F^2$, $G\subset A_5$, therefore $G$ is conjugate to $\mathrm{AGL}(1,{𝔽}_5)\cap A_5$, or $⟨(12345)⟩$.

By theorem 13.2.6 (c), $G$ is conjugate to $⟨(12345)⟩$ if and only if $f$ splits completely over $F(\alpha )$, and this gives the two last lines of the table.